Assistance With The Scanner

[TheUncertainyAnswer]

So i'm trying to make a basic text based resource game. I'm using the console scanner as the means of user input. I've added (or rather am trying to add) some manner of error checking to evaluate if the scanner input is invalid meaning the input is either a string or an integer that is greater than what is permitted (in my code the int is called t_available).

while(!scanner.hasNextInt() || scanner.nextInt() > t_available){
System.out.println("Invalid Selection");
System.out.println("Currently You Have Materials To Build " + h_temp + " Hovels");
System.out.println("And " + t_available + " Tradesmen Available");
System.out.println("How Many Hovels Would You Like To Build?");
scanner.next();
}

h_temp2 = scanner.nextInt();

If I remove the (scanner.nextInt() > t_available) portion it works as expected, but so long as its in there it doesn't appear the while loop breaks if a valid value (meaning < t_available) in input. It does flag an invalid value however (meaning > t_available). Can't figure this out for the life of me. Is there a better way to approach this? Any insight would be greatly appreciated.

[Norm]

scanner.nextInt() > t_available

That statement reads an int value, compares its value and throws the value away. Is that what you intend the code to do? If you want to save the value read, you need to read it into a variable.

[TheUncertainyAnswer]

Hi Norm, thanks for giving this a look. The idea is if the scanner value is invalid (a string or greater than t_available) you should be prompted to enter a valid value. If the value is invalid it should be thrown away, but if it is valid the while loop shouldn't activate and the code should proceed to the next line (h_temp2 = scanner.nextInt();). Does that make sense? Thanks in advance.

[Norm]

It seems confusing to have the user enter two values and only use the second one if the first one is valid.